how can we check association b/w an ordinal and categorical variable with four categories

[kusum shekhawat]

how can we check relationship b/w an ordinal and categorical variable with 4 categories? i have a variable with satisfaction score from 1-5, and other variable is distance from home like 1) <5km 2) 5-10km 3)10-15km 4) >15km

1. chi square is for categorical
2. spearman rank correlation is for ordinal, can someone help me with this?

[Carlo Lazzaro]

Kusum:
the -ologit- entry, Stata .pdf manual suggest something along the following lines:

. use "C:\Program Files\Stata17\ado\base\a\auto.dta"
(1978 automobile data)

. tabulate rep78 foreign, chi2

    Repair |
    record |      Car origin
      1978 |  Domestic    Foreign |     Total
-----------+----------------------+----------
         1 |         2          0 |         2 
         2 |         8          0 |         8 
         3 |        27          3 |        30 
         4 |         9          9 |        18 
         5 |         2          9 |        11 
-----------+----------------------+----------
     Total |        48         21 |        69 

          Pearson chi2(4) =  27.2640   Pr = 0.000

. ologit rep78 i.foreign

Iteration 0:   log likelihood = -93.692061  
Iteration 1:   log likelihood = -79.696089  
Iteration 2:   log likelihood = -79.034005  
Iteration 3:   log likelihood = -79.029244  
Iteration 4:   log likelihood = -79.029243  

Ordered logistic regression                             Number of obs =     69
                                                        LR chi2(1)    =  29.33
                                                        Prob > chi2   = 0.0000
Log likelihood = -79.029243                             Pseudo R2     = 0.1565

------------------------------------------------------------------------------
       rep78 | Coefficient  Std. err.      z    P>|z|     [95% conf. interval]
-------------+----------------------------------------------------------------
     foreign |
    Foreign  |    2.98155   .6203644     4.81   0.000     1.765658    4.197442
-------------+----------------------------------------------------------------
       /cut1 |  -3.158382   .7224269                     -4.574313   -1.742452
       /cut2 |  -1.362642   .3557343                     -2.059868   -.6654154
       /cut3 |   1.232161   .3431227                      .5596532     1.90467
       /cut4 |   3.246209   .5556657                      2.157124    4.335293
------------------------------------------------------------------------------

.

[kusum shekhawat]

thanks for the quick response Carlo Lazzaro, i have included my output, but for chi^2, p value is significant but for ologit it's not, what could be the possible reason for that?

 ta  Q18 Q13, chi

   18. How |
 satisfied |
   are you |
 with your |
 treatment |
        at |
  
Mohalla |
 Clinic on |
a scale of |   13. How far is the Mohalla 
Clinic from your home?
         1 |    1-2 km     2-5 km    5-10 km     5-10km      <1 km |     Total
-----------+-------------------------------------------------------+----------
         1 |         0          0          1          1          0 |         2 
         2 |         4         10          0          0          3 |        17 
         3 |        46         60          3         11         17 |       137 
         4 |        66         42          2          8         24 |       142 
         5 |        47         35          3          4         13 |       102 
-----------+-------------------------------------------------------+----------
     Total |       163        147          9         24         57 |       400 

         Pearson chi2(16) =  44.7348   Pr = 0.000

. ologit Q18 i.Q13

Iteration 0:   log likelihood = -497.52284  
Iteration 1:   log likelihood = -493.23756  
Iteration 2:   log likelihood =  -493.2324  
Iteration 3:   log likelihood =  -493.2324  

Ordered logistic regression                             Number of obs =    400
                                                        LR chi2(4)    =   8.58
                                                        Prob > chi2   = 0.0725
Log likelihood = -493.2324                              Pseudo R2     = 0.0086

------------------------------------------------------------------------------
         Q18 | Coefficient  Std. err.      z    P>|z|     [95% conf. interval]
-------------+----------------------------------------------------------------
         Q13 |
     2-5 km  |  -.5607657   .2107533    -2.66   0.008    -.9738346   -.1476969
    5-10 km  |  -.3627827   .6804225    -0.53   0.594    -1.696386    .9708208
     5-10km  |  -.7165465   .3989188    -1.80   0.072    -1.498413    .0653199
      <1 km  |  -.2549347     .27705    -0.92   0.357    -.7979427    .2880733
-------------+----------------------------------------------------------------
       /cut1 |  -5.621641   .7200313                     -7.032877   -4.210406
       /cut2 |  -3.322487   .2659922                     -3.843822   -2.801152
       /cut3 |  -.7400257   .1517087                     -1.037369   -.4426822
       /cut4 |   .8042331   .1530726                      .5042163     1.10425
------------------------------------------------------------------------------

[kusum shekhawat]

and if I'm not wrong, one of the assumption for chi square is that each cell must contain atleast 5 observations

[Carlo Lazzaro]

Kusum:
1) what if you go -testparm i.Q13-?;
2) correct. You can go -Fisher- exact test:

. tabulate rep78 foreign, exact

Enumerating sample-space combinations:
stage 5:  enumerations = 1
stage 4:  enumerations = 3
stage 3:  enumerations = 24
stage 2:  enumerations = 203
stage 1:  enumerations = 0

    Repair |
    record |      Car origin
      1978 |  Domestic    Foreign |     Total
-----------+----------------------+----------
         1 |         2          0 |         2 
         2 |         8          0 |         8 
         3 |        27          3 |        30 
         4 |         9          9 |        18 
         5 |         2          9 |        11 
-----------+----------------------+----------
     Total |        48         21 |        69 

           Fisher's exact =                 0.000

.

[Joro Kolev]

The chi2 test of association is the most general way how test whether the levels of two variables are independent.

[Nick Cox]

There are, it seems, some major problems here -- that fortunately are easily fixed.

First, "< 1 km" is manifestly out of order.

Second, it seems likely that there is a spurious distinction between "5-10 km" and "5-10km".

My guess is that these problems arise from a careless encode. That command won't protect you from errors.

Second, a common criterion for chi-square tests is that expected frequencies be at least 5, although in principle and practice frequencies with at least 1 seems to work quite well.

After surgery the data look like this:

* Example generated by -dataex-. For more info, type help dataex
clear
input byte(score dist freq)
1 1  0
1 2  0
1 3  0
1 4  2
2 1  3
2 2  4
2 3 10
2 4  0
3 1 17
3 2 46
3 3 60
3 4 14
4 1 24
4 2 66
4 3 42
4 4 10
5 1 13
5 2 47
5 3 35
5 4  7
end
label values dist dist
label def dist 1 "<1 km", modify
label def dist 2 "1-2 km", modify
label def dist 3 "2-5 km", modify
label def dist 4 "5-10 km", modify

I ran tabchi from tab_chi on SSC and floatplot from SSC.

tabchi score dist [w=freq]

(frequency weights assumed)

          observed frequency
          expected frequency

----------------------------------------------
          |                dist              
    score |   <1 km   1-2 km   2-5 km  5-10 km
----------+-----------------------------------
        1 |       0        0        0        2
          |   0.285    0.815    0.735    0.165
          |
        2 |       3        4       10        0
          |   2.422    6.928    6.247    1.403
          |
        3 |      17       46       60       14
          |  19.523   55.828   50.347   11.303
          |
        4 |      24       66       42       10
          |  20.235   57.865   52.185   11.715
          |
        5 |      13       47       35        7
          |  14.535   41.565   37.485    8.415
----------------------------------------------

6 cells with expected frequency < 5
4 cells with expected frequency < 1

          Pearson chi2(12) =  37.1824   Pr = 0.000
 likelihood-ratio chi2(12) =  26.2208   Pr = 0.010

A more cautious approach would merge scores 1 and 2, but that looks convincing to me as indicating an association.

The real question is what is the nature of that association. A graph never does any harm.

label var dist "distance to clinic"

floatplot score [w=freq], over(dist) fcolors(red*0.6 red*0.2 blue*0.2 blue*0.6 blue) centre(3) vertical lcolors(red red blue blue blue)

Nothing rules out a logit fit too. More serious is what other mess is to be found if encode was applied carelessly to other variables.

[Nick Cox]

FWIW. especially if you live and die by P < 0.05, the ordinal logit is better once the data are fixed

I am tempted to see fine structure in the graph that the model won't catch so well. For example, is there a notable shift at around 2 km? Perhaps many people are willing to walk or cycle 2 km, but beyond that is a bigger deal. Depends a great deal on who is being surveyed, and how mobile they are in all senses.

[kusum shekhawat]

thank you Nick Cox for the identifying the mistake i made using "encode". i'll surely check for other variables as well
but as you said score 1&2 can be merged, but what if i don't want to merge these satisfaction score
can i still use chi square?
and for fisher exact, some memory exceeded error is coming

. ta  Q18 Q13, exact

Enumerating sample-space combinations:
stage 5:  enumerations = 1
stage 4:  enumerations = 10
stage 3:  enumerations = 5038
stage 2:
exceeded memory limits using exact(1); try again with larger #; see help tabulate for details

   18. How |
 satisfied |
   are you |
 with your |
 treatment |
        at |
  
Mohalla |
 Clinic on |  RECODE of Q13 (13. How far is the Mohalla
a scale of |          
Clinic from your home?)
         1 |      <1km      1-2km      2-5km     5-10km |     Total
-----------+--------------------------------------------+----------
         1 |         0          0          0          2 |         2 
         2 |         3          4         10          0 |        17 
         3 |        17         46         60         14 |       137 
         4 |        24         66         42         10 |       142 
         5 |        13         47         35          7 |       102 
-----------+--------------------------------------------+----------
     Total |        57        163        147         33 |       400 

r(910);

[kusum shekhawat]

i tried the combination of these Independent variables. hoping I'm doing it correct now

 tab1 Q18 Q13 Q14 Q15 Q16 Q17

-> tabulation of Q18

    18. How |
  satisfied |
    are you |
  with your |
  treatment |
at 
Mohalla |
Clinic on a |
 scale of 1 |      Freq.     Percent        Cum.
------------+-----------------------------------
          1 |          2        0.50        0.50
          2 |         17        4.25        4.75
          3 |        137       34.25       39.00
          4 |        142       35.50       74.50
          5 |        102       25.50      100.00
------------+-----------------------------------
      Total |        400      100.00


-> tabulation of Q13  

  RECODE of |
   Q13 (13. |
 How far is |
the Mohalla |
    
Clinic |
  from your |
     home?) |      Freq.     Percent        Cum.
------------+-----------------------------------
       <1km |         57       14.25       14.25
      1-2km |        163       40.75       55.00
      2-5km |        147       36.75       91.75
     5-10km |         33        8.25      100.00
------------+-----------------------------------
      Total |        400      100.00

-> tabulation of Q14  

         14. How |
   frequently do |
      you go for |
 consulting 
the |
 doctor for your |
 diabetes treatm |      Freq.     Percent        Cum.
-----------------+-----------------------------------
 every 7-15 days |        112       28.00       28.00
     every month |        201       50.25       78.25
once in 3 months |         87       21.75      100.00
-----------------+-----------------------------------
           Total |        400      100.00

-> tabulation of Q15  

 15. Do you |
    need to |
     take a |
      prior |
appointment |
 
or do you |
walk-in for |
consultatio |
         n? |      Freq.     Percent        Cum.
------------+-----------------------------------
appointment |         88       22.00       22.00
    walk-in |        312       78.00      100.00
------------+-----------------------------------
      Total |        400      100.00

-> tabulation of Q16  

  RECODE of |
Q16 (16. On |
an average, |
how long do |
you need to |
    wait to |
  
see your |
       doct |      Freq.     Percent        Cum.
------------+-----------------------------------
   <15 mins |        183       45.75       45.75
 15-30 mins |        206       51.50       97.25
 30mins-1hr |         11        2.75      100.00
------------+-----------------------------------
      Total |        400      100.00

-> tabulation of Q17  

  RECODE of |
Q17 (17. On |
 an average |
   how much |
  time does |
   he spend |
       
for |
consultatio |
        n f |      Freq.     Percent        Cum.
------------+-----------------------------------
    <5 mins |         82       20.50       20.50
   5-10mins |        117       29.25       49.75
  10-20mins |        156       39.00       88.75
   >20 mins |         45       11.25      100.00
------------+-----------------------------------
      Total |        400      100.00

. ologit Q18 i.Q13 i.Q14 i.Q15 i.Q16 i.Q17

Iteration 0:   log likelihood = -497.52284  
Iteration 1:   log likelihood = -481.67219  
Iteration 2:   log likelihood = -481.59542  
Iteration 3:   log likelihood = -481.59538  

Ordered logistic regression                             Number of obs =    400
                                                        LR chi2(11)   =  31.85
                                                        Prob > chi2   = 0.0008
Log likelihood = -481.59538                             Pseudo R2     = 0.0320

-----------------------------------------------------------------------------------
              Q18 | Coefficient  Std. err.      z    P>|z|     [95% conf. interval]
------------------+----------------------------------------------------------------
              Q13 |
           1-2km  |   .1822775   .2836583     0.64   0.520    -.3736825    .7382375
           2-5km  |  -.0989544   .3377742    -0.29   0.770    -.7609798    .5630709
          5-10km  |  -.0957457   .4523877    -0.21   0.832    -.9824094    .7909179
                  |
              Q14 |
     every month  |   .4912631   .2504297     1.96   0.050     .0004299    .9820964
once in 3 months  |   1.072121   .3078421     3.48   0.000     .4687613     1.67548
                  |
              Q15 |
         walk-in  |  -.5169155   .2555246    -2.02   0.043    -1.017735   -.0160964
                  |
              Q16 |
      15-30 mins  |  -.0819063    .192771    -0.42   0.671    -.4597306    .2959179
      30mins-1hr  |  -.2450428   .7186659    -0.34   0.733    -1.653602    1.163517
                  |
              Q17 |
        5-10mins  |   .3171456   .3323564     0.95   0.340     -.334261    .9685523
       10-20mins  |   .4906557      .2606     1.88   0.060    -.0201109    1.001422
        >20 mins  |  -.1959378   .3984827    -0.49   0.623    -.9769495    .5850739
------------------+----------------------------------------------------------------
            /cut1 |  -5.114562   .8446824                     -6.770109   -3.459015
            /cut2 |  -2.800087   .5135582                     -3.806642   -1.793531
            /cut3 |  -.1355627   .4683834                     -1.053577    .7824519
            /cut4 |   1.478066   .4756921                      .5457264    2.410405
-----------------------------------------------------------------------------------

.

[Nick Cox]

The deal about low expected frequencies is to check that a low P-value isn’t just a side-effect but that you get consistent results both with and without pooling cells.

The problem with throwing many predictors into your model is coping with relationships between them, such as between time and distance.

[John Kane]

how can we check relationship b/w an ordinal and categorical variable with 4 categories? i have a variable with satisfaction score from 1-5, and other variable is distance from home like 1) <5km 2) 5-10km 3)10-15km 4) >15km

1. chi square is for categorical
2. spearman rank correlation is for ordinal, can someone help me with this?

Personally, I would recommend a chi-squared test if your only interest is in a bivariate relationship. The ordered logit model involves the assumption of proportional odds, whereas the chi-squared test makes no such assumption and will effectively regard each value of each variable as its own category. Its results (i.e., the cross-tab) can also be far more informative than the single coefficient you will get from an ologit model.

If you want to add controls, then that's a different story, but if it's just a bivariate relationship that you are interested in, the chi-squared test is the safest and most useful, in my view. (p.s., both of the variables you describe sound ordinal to me; neither is purely categorical as each has a natural ordering of values.)

[Joseph Coveney]

both of the variables you describe sound ordinal to me; neither is purely categorical as each has a natural ordering of values.

To me, too. If the OP is interested in assessing the degree of association between these two ordered-categorical variables, then consider -tabulate- with its two measures of ordinal association, as illustrated below (using Nick's cleaned up dataset from #7 above).

OP can also fit an ordered-categorical regression model with the two variables, especially, if the assumption of exogeneity of distance is tenable; even so, -contrast- can give a measure of association in its linear component of the set of orthogonal polynomial contrasts (also illustrated below). With both envisioned as purely outcome variables, then perhaps something like a bivariate ordered-probit model could be helpful.

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------------------------------------------------------------------------------
ÿÿÿÿÿÿÿÿÿsatÿ|ÿCoefficientÿÿStd.ÿerr.ÿÿÿÿÿÿzÿÿÿÿP>|z|ÿÿÿÿÿ[95%ÿconf.ÿinterval]
-------------+----------------------------------------------------------------
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-------------+----------------------------------------------------------------
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------------------------------------------------------------------------------

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------------------------------------------------
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-------------+----------------------------------
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ÿÿÿ(linear)ÿÿ|ÿÿÿÿÿÿÿÿÿÿ1ÿÿÿÿÿÿÿÿ4.54ÿÿÿÿÿ0.0330
(quadratic)ÿÿ|ÿÿÿÿÿÿÿÿÿÿ1ÿÿÿÿÿÿÿÿ1.57ÿÿÿÿÿ0.2096
ÿÿÿÿ(cubic)ÿÿ|ÿÿÿÿÿÿÿÿÿÿ1ÿÿÿÿÿÿÿÿ2.70ÿÿÿÿÿ0.1006
ÿÿÿÿÿÿJointÿÿ|ÿÿÿÿÿÿÿÿÿÿ3ÿÿÿÿÿÿÿÿ8.70ÿÿÿÿÿ0.0336
------------------------------------------------

.ÿ
.ÿ//ÿOtherwise,ÿmaybeÿconsiderÿsomethingÿalongÿtheÿfollowingÿlines
.ÿquietlyÿdropÿifÿ!count

.ÿquietlyÿexpandÿcount

.ÿgsemÿ(sat@1ÿ<-ÿF1,ÿoprobit)ÿ(dis@1ÿ<-ÿF2,ÿoprobit),ÿvariance(F1@1ÿF2@1)ÿ///
>ÿÿÿÿÿcovariance(F1*F2)ÿnocnsreportÿnoheaderÿnolog
------------------------------------------------------------------------------
ÿÿÿÿÿÿÿÿÿÿÿÿÿ|ÿCoefficientÿÿStd.ÿerr.ÿÿÿÿÿÿzÿÿÿÿP>|z|ÿÿÿÿÿ[95%ÿconf.ÿinterval]
-------------+----------------------------------------------------------------
satÿÿÿÿÿÿÿÿÿÿ|
ÿÿÿÿÿÿÿÿÿÿF1ÿ|ÿÿÿÿÿÿÿÿÿÿ1ÿÿ(constrained)
-------------+----------------------------------------------------------------
disÿÿÿÿÿÿÿÿÿÿ|
ÿÿÿÿÿÿÿÿÿÿF2ÿ|ÿÿÿÿÿÿÿÿÿÿ1ÿÿ(constrained)
-------------+----------------------------------------------------------------
/satÿÿÿÿÿÿÿÿÿ|
ÿÿÿÿÿÿÿÿcut1ÿ|ÿÿ-3.663671ÿÿÿ.3513755ÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿ-4.352354ÿÿÿ-2.974988
ÿÿÿÿÿÿÿÿcut2ÿ|ÿÿ-2.361907ÿÿÿ.1518846ÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿ-2.659595ÿÿÿ-2.064219
ÿÿÿÿÿÿÿÿcut3ÿ|ÿÿ-.3927769ÿÿÿ.0899474ÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿ-.5690705ÿÿÿ-.2164834
ÿÿÿÿÿÿÿÿcut4ÿ|ÿÿÿ.9330804ÿÿÿ.0958584ÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿ.7452014ÿÿÿÿ1.120959
-------------+----------------------------------------------------------------
/disÿÿÿÿÿÿÿÿÿ|
ÿÿÿÿÿÿÿÿcut1ÿ|ÿÿ-1.512085ÿÿÿ.1095182ÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿ-1.726736ÿÿÿ-1.297433
ÿÿÿÿÿÿÿÿcut2ÿ|ÿÿÿ.1757111ÿÿÿ.0889448ÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿ.0013826ÿÿÿÿ.3500396
ÿÿÿÿÿÿÿÿcut3ÿ|ÿÿÿÿ1.96634ÿÿÿ.1278867ÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿ1.715687ÿÿÿÿ2.216994
-------------+----------------------------------------------------------------
ÿÿÿÿÿÿvar(F1)|ÿÿÿÿÿÿÿÿÿÿ1ÿÿ(constrained)
ÿÿÿÿÿÿvar(F2)|ÿÿÿÿÿÿÿÿÿÿ1ÿÿ(constrained)
-------------+----------------------------------------------------------------
ÿÿÿcov(F1,F2)|ÿÿ-.2389781ÿÿÿ.1140519ÿÿÿÿ-2.10ÿÿÿ0.036ÿÿÿÿ-.4625157ÿÿÿ-.0154405
------------------------------------------------------------------------------

.ÿassertÿe(converged)

.ÿ
.ÿexit

endÿofÿdo-file


.